The lenght of rectifiable curve in $\mathbb{R}^n \setminus B[0,r]$ that connects antipodes points.

Question

This question is from my homework, here it goes:

Let $\gamma \colon [a,b] \to \mathbb{R}^n \setminus B[0,r]$ be a rectifiable curve such that $\gamma(a)=-\gamma(b)$. Using euclidean norm prove that $L(\gamma)>\pi r$.

My attempt: Define $\alpha \colon [a,b] \to \mathbb{R}^n$, $\alpha(t)=\frac{\gamma(t)}{||\gamma(t)||}r$. Using uniform continuity ,since $[a,b]$ is compact, $\exists$ $\delta>0$ such that $|x-y|< \delta \implies ||\gamma(x) - \gamma(y)|< r$.

Now, for any partition $P=\{a=t_0<\ldots<t_n\}$ such that $|P|< \delta$, I can prove that $\langle \gamma(t_i) , \gamma(t_{i-1})\rangle > 0$.

So, I want prove that $||\gamma(t_i) - \gamma(t_{i-1})||^2 > ||\alpha(t_i)-\alpha(t_{i-1})||^2$ using that inner product, and conlcude that:

$L(\gamma, P)= \displaystyle \sum_{i=1}^{n}||\gamma(t_i) - \gamma(t_{i-1})|| > \displaystyle \sum_{i=1}^{n}||\alpha(t_i) - \alpha(t_{i-1})|| = L(\alpha,P)$. And after that $L(\alpha) \geq \pi r$.

Unfortunately I could not prove that $||\gamma(t_i) - \gamma(t_{i-1})||^2 > ||\alpha(t_i)-\alpha(t_{i-1})||^2$, and also don't know how to prove that $L(\alpha)\geq \pi r$.

Thank you for any help.

Answer 1

Given $u,v$ with $\|u\|\ge\|v\|\ge r>0$ we have $\|u-v\|^2\ge \left\|\frac{\|v\|}{\|u\|}u-v\right\|^2$. This follows from $$ \begin{align}\left\langle u-\frac{\|v\|}{\|u\|}u,\frac{\|v\|}{\|u\|}u-v\right\rangle&=\|v\|\|u\|-\langle u,v\rangle-\|v\|^2+\frac{\|v\|}{\|u\|}\langle u,v\rangle\\&=\left(1-\frac{\|v\|}{\|u\|}\right)(\|v\|\|u\|-\langle u,v\rangle)\\&\ge 0\end{align}$$ and therefore $$\begin{align}\left\|u-v\right\|^2 &=\left\|u-\frac{\|v\|}{\|u\|}u\right\|^2+\left\|\frac{\|v\|}{\|u\|}u-v\right\|^2+2\left\langle u-\frac{\|v\|}{\|u\|}u,\frac{\|v\|}{\|u\|}u-v\right\rangle\\&\ge \left\|\frac{\|v\|}{\|u\|}u-v\right\|^2\end{align}$$ Consequently, $$ \left\|\frac r{\|u\|}u-\frac r{\|v\|}v\right\|= \frac r{\|v\|}\left\|\frac {\|v\|}{\|u\|}u-v\right\|\le \frac r{\|v\|}\|u-v\|\le \|u-v\|$$ This is your $\gamma$-$\alpha$-inequality.

For the final inequality, you may want to compare $\alpha$ to a path following a single meridian (with the same "latitudes").